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Blababa [14]
3 years ago
10

A group of 25 particles have the following speeds: two have speed 11m/s , seven have 16m/s , four have 22m/s , three have 25m/s

, six have 31m/s , one has 34m/s , and two have 41 m/s .
a. Determine the average speed.
vavg= ______________m/s
b. Determine the rms speed.
vrms= ______________m/s
c. Determine the most probable speed.
Physics
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

a) vavg = 16.59 m/s

b) vrms= 26.42 m/s

c) the most probable speed is the average speed = 16.59 m/s  (since the is no particle with 16.59 m/s → 16m/s is the most probable speed observed)

Explanation:

a) the average velocity will be

vavg =( 1/total number of particles )*∑ velocity * number of particles  

= (2*11 + 4*16 + 4*22 +3*25 + 6*31  + 1*34 + 2*41)/ (2 + 4 + 4 +3 + 6 + 1 + 2) = 16.59 m/s

b) the root mean squared speed (rms) will be

vrms=√[( 1/total number of particles )*∑ velocity² * number of particles] =   √[ (2*11² + 4*16² + 4*22² +3*25² + 6*31²  + 1*34² + 2*41²)/ (2 + 4 + 4 +3 + 6 + 1 + 2)]  = 26.42 m/s

c) if any particle is chosen at random then the probability of choosing that particle will be the given be the fraction of particles over the total , and thus the expected speed (most probable speed ) will be the average speed = 16.59 m/s ( since the is no particle with 16.59 m/s → 16m/s is the most probable speed observed)

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tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

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c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

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There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

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r= R_{E}cos \theta

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r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

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- If the magnetic flux through a coil is increasing, then the induced current has a direction such that the magnetic field generated by the coil is opposite to the direction of the external magnetic field, in order to decrease the total flux

- If the magnetic flux through a coil is decreasing, then the induced current has a direction such that the magnetic field generated by the coil is in the same direction as the external field, in order to increase the total flux

2. False

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\Phi = BA cos \theta

where

B is the magnitude of the magnetic field

A is the surface area

\theta is the angle between the direction of B and the normal vector to the surface

As we see, the magnetic flux depends not only on B and A, but also on the orientation of the coil with respect to the magnetic field.

3. False

Explanation:

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